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A pendulum bob of mass m connected to th...

A pendulum bob of mass `m` connected to the end of material string of length `l` is released from rest from horizontal position as shown in the figure. At the lowest point the bob makes an elastic collision with a stationary block of mass `5m`, which is kept on a frictionless surface. Choose out the correct statement(s) for the instant just after the impact.

A

Tension in the string at lowest point just after collision is `(17//9) mg`

B

Tension in the string at lowest point just before collision is `3 mg`

C

The velocity of the block is `sqrt(2gl)//3`

D

The maximum height attained by the pendulum bob after impact is (measured from the lowest position) `(4l)/(9)`

Text Solution

Verified by Experts

The correct Answer is:
A, B, C, D
The velocity of bob just before the impact `v=sqrt(2gl)` along the horizontal direction

Before collision After Collision
From momentum conservation
`mv=mv_(1)+5mv_(2)`
From coefficient of restitution equation
`1=(v_(1)+v_(2))/(v)rArrv_(1)+v_(2)=v`
Solving above equations, we get , `v_(1)=(2v)/(3), v_(2)=(v)/(3)`
For tension in string , `T-mg=(mv_(1)^(2))/(l)rArrT=(17)/(9)mg`
`T-mg=(mv^(2))/(l), V_(2)=(sqrt(2gl))/(3)(T=3mg)`
(i) Let the maximum height attained by the bob be `h`, then `(mv_(1)^(2))/(2)=mghrArrh=(4l)/(9)`
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